Peaceable Armies of Queens: Difference between revisions

I found this challenging constraint programming problem in the following paper:

• B. M. Smith, K. E. Petrie, and I. P. Gent. Models and symmetry breaking for peaceable armies of queens. In Integration of AI and OR Techniques in Constraint Programming for Combinatorial Optimization Problems, pages 271–286. Springer Berlin Heidelberg, 2004.

A variation of the problem can be found on page 329 of the Handbook on Constraint programming.

The challenge is to place two equal-sized armies of white and black queens onto a chessboard. We can distinguish between two problems:

• checking whether two armies of n queens can be placed on a dim*dim chessboard,
• for a give board of size dim*dim find the maximal size n of armies that can be placed onto the board, i.e., solving an optimisation problem.

Checking

Here is a B model for the checking problem. The problem is quite straightforward to encode in B. Many lines below (in the DEFINITIONS) just define the animation function to graphically represent solutions.

MACHINE JustQueens
DEFINITIONS
      SET_PREF_TIME_OUT == 1000000;
      SET_PREF_MAX_INITIALISATIONS == 1;
      ANIMATION_IMG0 == "images/sm_empty_box.gif";
      ANIMATION_IMG1 == "images/sm_gray_box.gif";
      ANIMATION_IMG2 == "images/sm_queen_white.gif";
      ANIMATION_IMG3 == "images/sm_queen_black.gif";
      ANIMATION_IMG4 == "images/sm_knight_white.gif";
      ANIMATION_IMG5 == "images/sm_knight_black.gif";
      ANIMATION_IMG6 == "images/sm_white_queen_white.gif";
      ANIMATION_IMG7 == "images/sm_white_queen_black.gif";
      BWOFFSET(xx,yy) == (xx+yy) mod 2;
      ANIMATION_FUNCTION_DEFAULT == ( {r,c,i|r:1..dim & c:1..dim & i=(r+c) mod 2 }  );
      ANIMATION_FUNCTION == ( UNION(k).(k:1..n| {(blackc(k),blackr(k),2+BWOFFSET(blackc(k),blackr(k)))}) \/
                              UNION(k).(k:1..n| {(whitec(k),whiter(k),6+BWOFFSET(whitec(k),whiter(k)))})    );
      ORDERED(c,r) == (!i.(i:1..(n-1) => c(i) <= c(i+1)) &
                       !i.(i:1..(n-1) => (c(i)=c(i+1) => r(i) < r(i+1))))
CONSTANTS n, dim, blackc, blackr, whitec, whiter
PROPERTIES
 n = 8 & dim = 8 &
 blackc : 1..n --> 1..dim &
 whitec : 1..n --> 1..dim &
 blackr : 1..n --> 1..dim &
 whiter : 1..n --> 1..dim &
 ORDERED(blackc,blackr) &  /* ensures proper ordering + that we do not place two queens on same square */
 ORDERED(whitec,whiter) &
 
 !(i,j).(i:1..n & j:1..n => blackc(i) /= whitec(j)) &
 !(i,j).(i:1..n & j:1..n => blackr(i) /= whiter(j)) &
 !(i,j).(i:1..n & j:1..n => blackr(i) /= whiter(j) + (blackc(i)-whitec(j))) &
 !(i,j).(i:1..n & j:1..n => blackr(i) /= whiter(j) - (blackc(i)-whitec(j))) &
 
  whitec(1) < blackc(1) /* symmetry breaking */
END

Here are some running times on my MacBook Air 1.7 GHz i7, also comparing to using [Using_ProB_with_KODKOD Kodkod as ProB's backend] and using an [Alloy] model (see below).

 dim=7, n=7 : solved in 0.3 secs
 dim=7, n=8 : unsat in 20 secs
 dim=8, n=6 : solved in 0.02 secs   (1.12 secs with Kodkod)
 dim=8, n=7 : solved in 0.06 secs   (2.66 secs with Kodkod)
 dim=8, n=8 : solved in 0.53 secs   (8.44 secs with Kodkod; 7.03 secs second run; +/- 8.5 with Alloy & miniSat; 9.3 seconds if we avoid overflows)
 dim=8, n=9 : solved in 12.96 secs  (128.07 secs with Kodkod ; +/- 84 secs with Alloy & miniSat)
 dim=8, n=10 : unsat in 728 secs    (Alloy & miniSat still running after over four hours)

The first solution found for dim=8 and n=9 is as follows:

The Alloy model for comparison

Here is the [Alloy] model we have used. Initially, the model gave us wrong solutions as we were using + instead of sum; this was correct for previous versions of Alloy but not for Alloy 4.2 which we were using.

abstract sig Queens {
  row : one Int,
  col: one Int,
} {
 row >= 0 and row < 8
 and col >= 0 and col < 8
}
sig BQueens extends Queens {} {}
sig WQueens extends Queens {} {}

pred nothreat(q1,q2 : Queens) {
q1.row != q2.row
and q1.col != q2.col
and plus[ int[q1.row] , int[q1.col]]  != plus[ int[q2.col] ,  int[q2.row]]
    and plus [int[q1.row] , int[q2.col]] != plus[ int[q1.col]  , int[q2.row]]
}

pred nothreats { all q1:BQueens, q2:WQueens |
    nothreat[q1, q2]
}

pred alldiffB { all q1:BQueens, q2:BQueens |
  q1=q2 or q1.row != q2.row or q1.col != q2.col
}
pred alldiffW { all q1:WQueens, q2:WQueens |
  q1=q2 or q1.row != q2.row or q1.col != q2.col
}

pred equalnum {
    #(WQueens) = #(BQueens)
}

pred valid {
  nothreats and equalnum and alldiffB and alldiffW
}

fact {
  #Queens =  16
}

run valid for 16 Queens, 7 int

Optimization

One can obviously use the above models to try and manually find a maximal value of n for a given board size dim. Below, we have encoded this process as B constraints, by setting up the problem twice: once for the army size n and a second time for army size n+1. The second encoding is wrapped into a negated existential quantification. (This model can no longer be translated into Kodkod because of this.)

ProB solves this optimisation problem in about 780 seconds for a board size of 8.

MACHINE JustQueens_FindOptimal_CBC
/*
 dim=5  -->  optimum n=4 found in 0.18 secs
 dim=6  -->  optimum n=5 found in 1.16 secs
 dim=7  -->  optimum n=7 found in 21.174 secs
 dim=8  -->  optimum n=9 found in 780.130 secs
*/

DEFINITIONS
      SET_PREF_TIME_OUT == 1000000;
      SET_PREF_MAX_INITIALISATIONS == 1;
      ANIMATION_IMG0 == "images/sm_empty_box.gif";
      ANIMATION_IMG1 == "images/sm_gray_box.gif";
      ANIMATION_IMG2 == "images/sm_queen_white.gif";
      ANIMATION_IMG3 == "images/sm_queen_black.gif";
      ANIMATION_IMG4 == "images/sm_knight_white.gif";
      ANIMATION_IMG5 == "images/sm_knight_black.gif";
      ANIMATION_IMG6 == "images/sm_white_queen_white.gif";
      ANIMATION_IMG7 == "images/sm_white_queen_black.gif";
      BWOFFSET(xx,yy) == (xx+yy) mod 2;
      ANIMATION_FUNCTION_DEFAULT == ( {r,c,i|r:1..dim & c:1..dim & i=(r+c) mod 2 }  );
      ANIMATION_FUNCTION == ( UNION(k).(k:1..n| {(blackc(k),blackr(k),2+BWOFFSET(blackc(k),blackr(k)))}) \/
                              UNION(k).(k:1..n| {(whitec(k),whiter(k),6+BWOFFSET(whitec(k),whiter(k)))})    );
      ORDERED(c,r,nn) == (!i.(i:1..(nn-1) => c(i) <= c(i+1)) &
                          !i.(i:1..(nn-1) => (c(i)=c(i+1) => r(i) < r(i+1))));
      CHECK_TYPE(bc,br,wc,wr,nn) == (
			 bc : 1..nn --> 1..dim &
			 wc : 1..nn --> 1..dim &
			 br : 1..nn --> 1..dim &
			 wr : 1..nn --> 1..dim );
      CHECK_DIAGONALS(bc,br,wc,wr,nn) == (
			 !(i,j).(i:1..nn & j:1..nn => bc(i) /= wc(j)) &
			 !(i,j).(i:1..nn & j:1..nn => br(i) /= wr(j)) &
			 !(i,j).(i:1..nn & j:1..nn => br(i) /= wr(j) + (bc(i)-wc(j))) &
			 !(i,j).(i:1..nn & j:1..nn => br(i) /= wr(j) - (bc(i)-wc(j)))
           )
CONSTANTS n, dim, blackc, blackr, whitec, whiter
PROPERTIES
 n : 1..16 & dim = 8 &
 
 CHECK_TYPE(blackc, blackr, whitec, whiter, n) &
 ORDERED(blackc,blackr,n) &  /* ensures proper ordering + that we do not place two queens on same square */
 ORDERED(whitec,whiter,n) &
 CHECK_DIAGONALS(blackc, blackr, whitec, whiter, n) &
 whitec(1) < blackc(1) /* symmetry breaking */ &

 /* Repeat constraints for n+1 and assert that it cannot be solved */
  not( #(n1,blackc1, blackr1, whitec1, whiter1). 
        (n1=n+1 & /* n1:2..17 & */
         CHECK_TYPE(blackc1, blackr1, whitec1, whiter1, n1) &
		 ORDERED(blackc1,blackr1,n1) &  /* ensures proper ordering + that we do not place two queens on same square */
		 ORDERED(whitec1,whiter1,n1) &
         CHECK_DIAGONALS(blackc1, blackr1, whitec1, whiter1, n1) &
		 whitec1(1) < blackc1(1) /* symmetry breaking */
      )
     )
END

Here are the solving times for various board sizes on my MacBook Air:

 dim=5  -->  optimum n=4 found in 0.18 secs
 dim=6  -->  optimum n=5 found in 1.16 secs
 dim=7  -->  optimum n=7 found in 21.174 secs
 dim=8  -->  optimum n=9 found in 780.130 secs

The first solution found for dim=8 is as follows: